Correlation and Regression
Domain 2: Correlation and Regression
This domain tests the direction and strength of linear association between two variables and the two regression equations that let you predict one from the other. Expect one numerical question (r from a table, or r and the lines from summary sums) plus a theory part on properties.
Core rules
- Pearson's coefficient: r = Σ(x − x̄)(y − ȳ)/√[Σ(x − x̄)²·Σ(y − ȳ)²] = Σxy/(n σx σy) using deviations.
- r is unit-free, lies in [−1, 1], and is unchanged by a shift of origin or by scaling, except that a negative scale factor reverses its sign.
- Regression of y on x: y − ȳ = b_yx(x − x̄), with b_yx = r·σy/σx = Σxy/Σx².
- Regression of x on y: x − x̄ = b_xy(y − ȳ), with b_xy = r·σx/σy = Σxy/Σy².
- r = ±√(b_yx · b_xy); both coefficients carry the sign of r, and |r| ≤ 1 forces b_yx · b_xy ≤ 1.
- The two lines always intersect at (x̄, ȳ) and coincide only when |r| = 1.
- Spearman: ρ = 1 − 6Σd²/[n(n² − 1)]; for each tie of m ranks add (m³ − m)/12 to Σd².
Worked example
Five pairs (1, 2), (2, 4), (3, 5), (4, 4), (5, 6): x̄ = 3, ȳ = 4.2, Σ(x − x̄)(y − ȳ) = 8, Σ(x − x̄)² = 10, Σ(y − ȳ)² = 8.8. b_yx = 8/10 = 0.8, so the y-on-x line is y = 4.2 + 0.8(x − 3) = 0.8x + 1.8. b_xy = 8/8.8 = 0.909, hence r = √(0.8 × 0.909) = 0.853. Check directly: r = 8/√(10 × 8.8) = 8/9.381 = 0.853.
Traps
- Swapping the two lines: the coefficient attached to x belongs to the y-on-x equation.
- Concluding that r = 0 means independence - it only rules out linear association.
- Reading causation into a high r.
- Using Σ(x − x̄)² in the denominator of b_xy instead of Σ(y − ȳ)².
- Ignoring tied ranks, which distorts ρ.